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A and B are two points separated by a distance 5cm. Two charges 10 $μC$ and 20 $μC$ are placed at A and B. The resultant electric intensity at a point P outside the charges at a distance 5cm from 10 $μC$ is

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zero

9 × 10⁶ N/C away from 10 $μC$

56 × 10⁶ N/C towards 10 $μC$

54 × 10⁶N/Caway from 10 $μC$

answer is A.

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Detailed Solution

from principle of superposition

$E_{p} = E_{A} + E_{B}$

$E_{p} = 9 \times 10^{9} (\frac{q_{A}}{r_{A}^{2}} + \frac{q_{B}}{r_{B}^{2}})$

$E_{p} = 54 \times 10^{6} N / C (away from 10 μC)$

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